You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integrating Factor

Let y=fx be t...

Question

Let $y = f (x)$ be the curve satisfy the differential equation $2 x y d y = (x^{2} + y^{2} + 1) d x$ such that $f (1) = 0.$ Then the value of $f^{2} (2)$ is

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Given: $2 x y d y = (x^{2} + y^{2} + 1) d x$
$\Rightarrow 2 y \frac{d y}{d x} - \frac{y^{2}}{x} = x + \frac{1}{x}$
Let $y^{2} = t \Rightarrow 2 y \frac{d y}{d x} = \frac{d t}{d x}$
$∴ \frac{d t}{d x} - \frac{t}{x} = x + \frac{1}{x}$
$I . F . = \frac{1}{x}$
Now, the general solution:
$t \cdot \frac{1}{x} = \int \frac{1}{x} (x + \frac{1}{x}) d x$
$\Rightarrow \frac{y^{2}}{x} = x - \frac{1}{x} + C$
Since $f (1) = 0,$ therefore $C = 0$
$∴ y^{2} = x^{2} - 1$
For $x = 2 :$
$f^{2} (2) = y^{2} (2) = 2^{2} - 1 = 3$

Suggest Corrections

Similar questions

2 x y d y - (x^{2} + y^{2} + 1) d x = 0.

2 x y d y = x^{2} + y^{2} + 1 d x

C_{1}

2 x y \frac{d y}{d x} = y^{2} - x^{2},

x > 0.

C_{2}

\frac{2 x y}{x^{2} - y^{2}} = \frac{d y}{d x} .

(1, 1),

C_{1}

C_{2}

(x^{2} - y^{2}) d x + 2 x y d y = 0

(1, 1)

y = f (x)

R

f (1) = 1

f (x)

x f^{'} (x) = x^{2} + f (x) - 2

Join BYJU'S Learning Program